I have recently started tinkering with technology to see how things work. I am building a small box with a phone charger, flashlight and solar panel however I wanted to add a laser pen. After taking it apart enough, I can't understand how it gets power from the 2 batteries in it, here is the pen, where would I connect the red and black wires to power it from my battery?

\$\begingroup\$Connect it to the original battery terminals. One is the case.\$\endgroup\$
– TransistorAug 21 '16 at 17:27

\$\begingroup\$The problem is I had to tear apart half of it to get down to that part of it. Tried just placing the batteries inside the case and pressing it against it but it never worked\$\endgroup\$
– Andre QueenAug 21 '16 at 17:29

\$\begingroup\$One terminal of the pushbutton is connected to a metal tab. That is probably one contact point.\$\endgroup\$
– Whit3rdAug 22 '16 at 7:21

\$\begingroup\$The right tab of the first image is the negative connection, I just don't know where the positive goes! All help appreciated\$\endgroup\$
– Andre QueenAug 22 '16 at 7:23

\$\begingroup\$Just remember that you have no room for mistakes: sometimes powering it the wrong way destroys it on the first attempt. Even Maglite flashlights sometimes state explicitly that inverting the polarity immediately destroys the circuit.\$\endgroup\$
– FarOAug 22 '16 at 10:15

1 Answer
1

Similar to a flashlight, one battery contact is the outer metal case and the other contact is the center. A laser diode will be sensitive to polarity. If you still remember which way the batteries were inserted before the teardown, the tab at the end of the PCB is the inner contact for the center battery terminal (most likely positive I'd guess).

You do promise to be careful not to burn your eyes out with what is presumably a Class I laser device, right? Wikipedia Laser Saftey is required reading.

\$\begingroup\$Also, do not be tempted to replace the AAA battery with a larger one until you understand the laser diode - it will probably be relying on the low current capacity / high internal resistance of the battery.\$\endgroup\$
– pjc50Aug 22 '16 at 9:56

\$\begingroup\$Hmm that's what I assumed however after wiring it up nothing happens, am I safe to assume at this point I've broken the light? P.s. don't worry I'm not a child who would shone it in my eyes or anything like that\$\endgroup\$
– Andre QueenAug 22 '16 at 9:58